Constructing a Triangle from Two Vertices and the Symmedian Point
|
|
- Delilah King
- 5 years ago
- Views:
Transcription
1 Forum Geometricorum Volume 18 (2018) FORUM GEOM ISSN Constructing a Triangle from Two Vertices and the Symmedian Point Michel Bataille Abstract. Given three noncollinear points A, B, K, we propose a simple straightedge-and-compass construction of C such that K is the symmedian point of triangle ABC. Conditions for the construction to be possible and the number of solutions are studied in details. 1. Introduction The problem of constructing a triangle ABC given two vertices A, B and one of the familiar centers (centroid, incenter, orthocenter, circumcenter), which appears in Wernick s famous list published in 1982 ([6, 7]), is easily solved. The case when the selected center is the symmedian point K being not listed, this article aims at offering a simple construction of the third vertex C from A, B, K and at studying the conditions on the three given points for the existence of a solution. In what follows, we always assume that the point K does not lie on the line AB. 2. The construction The construction of a suitable point C will rest upon a not-very-well-known property of the symmedians, stated and proved in [1]. To keep this paper selfcontained, it is presented here with a different proof. Proposition 1. Let ABC be a triangle and let B be the reflection of B in A. The symmedian through A is the tangent at A to the circumcircle of ΔCAB. Figure 1 Publication Date: May 17, Communicating Editor: Paul Yiu.
2 10 M. Bataille Proof. Let M be the midpoint of BC and let s be the symmedian through A, that is, the reflection of the median m = AM in the internal bisector l of BAC (Figure 1). Let U be the center of the circle Γ through C, A, B and S the point of intersection other than A of l and Γ. Since l is the external bisector of CAB, we have SC = SB and it follows that US is perpendicular to B C, hence to m = AM. As a result, s is perpendicular to U 1 S, where U 1 is the reflection of U in l, hence to UA since UAU 1 S is a rhombus. Turning to our problem, consider the three points A, B, K and let B be the reflection of B in A and A be the reflection of A in B. Let γ A (resp. γ B ) denote the unique circle passing through B (resp. A ) and tangent to AK at A (resp. tangent to BK at B). The point K is the symmedian point of ΔABC if and only if the lines AK and BK are the symmedians through A and B, respectively. Proposition 1 shows that this will occur if and only if the circles γ A and γ B both pass through C. Thus, the construction of the circles γ A and γ B readily yields the desired third vertex C as a common point of these circles, provided that this common point exists. Clearly, our problem has at most two solutions (see Figure 2, where two solutions C 1 and C 2 are obtained). Figure 2. The discussion From the construction above, the number of solutions for C depends on the relative position of the circles γ A and γ B. The way this position is related to the location of the points A, B, K is described in the following result: Theorem 2. The number of solutions for C is 0, 1, or2, according as the sum KA + KB is greater than, equal to, or less than 2AB. Proof. Let U and V be the centers of γ A and γ B, and let O denote the midpoint of AB (Figure 2). In a suitable system of axes with origin at O, wehave A( c 2, 0), B( c 2, 0) (where c = AB) and we set K(m, n). Expressing that V is the point of intersection of the perpendicular bisector of BA (with equation x = c) and the perpendicular to BK at B (whose equation ( m 2) c ( ) x + ny = c 2 m c 2
3 Constructing a triangle from two vertices and the symmedian point 11 is easily obtained), we find V (c, v) where v = c ( c 2n 2 m). Similarly, U( c, u) where u = c ( c 2n 2 + m) and a short calculation yields UV 2 = c2 (m 2 +4n 2 ) n 2, UA 2 = c2 ((2m + c) 2 +4n 2 ) 16n 2, VB 2 = c2 ((2m c) 2 +4n 2 ) 16n 2. (1) The condition UV > VB UA always holds because UV 2 =4c 2 + v u 2 =4c 2 +VB 2 c2 4 +UA2 c2 4 2uv > V B2 +UA 2 2UA VB. Thus, the circles γ A and γ B are secant (resp. tangent) if and only if UV < UA+ VB(resp. UV = UA+ VB). Now, from (1), and with some algebra, we see that UV < UA+ VBis equivalent to 28n 2 +4m 2 c 2 < (4n 2 +4m 2 + c 2 ) 2 16c 2 m 2, which itself is equivalent to 4m 2 +28n 2 <c 2 or m 2 +12n 2 <c 2, and finally to m 2 +12n 2 <c 2. The latter condition means that K is interior to the ellipse E with foci A and B and major axis 2c. Clearly, the circles γ A and γ B are tangent if and only if K lies on E. Since E is the set of all points P such that PA + PB = 2c, the proof is complete. 4. Properties of the solutions First, we examine the case when the problem has two solutions C 1 and C 2 (Figure 2) and prove the following Proposition. If C 1 C 2 and K is the symmedian point of both ΔABC 1 and ΔABC 2, then the line C 1 C 2 is a median of each triangle. Proof. We just observe that the midpoint O of AB has the same power with respect to γ A and γ B (since OA OB = c2 4 = OB OA ) and deduce that O is on the radical axis C 1 C 2 of these two circles. In the case when the solution C is unique, the triangle ABC has a feature which is worth mentioning: Proposition 4. If K is the symmedian point of ΔABC for a unique point C, then CA 2 + CB 2 =2AB 2. In other words, the triangle is a C-root-mean-square triangle (see [2, 4] for numerous properties of such triangles). Proof. Again, O lies on the radical axis of γ A and γ B, hence on the common tangent to γ A and γ B at C. It follows that c2 4 = OC 2 ; but, CO being the length of the median from C, we also have 4CO 2 =2(CA 2 + CB 2 ) c 2 and a short calculation gives the desired relation.
4 12 M. Bataille Note in passing that when K traverses the ellipse E, the corresponding unique solution C traverses the circle with center O and radius c 2 (less the common points with the line AB, of course). 5. Further remarks To conclude, we draw some interesting corollaries of the results above. In an arbitrary triangle ABC, not only does the symmedian point K satisfy KA + KB 2AB, but it also satisfies the similar inequalities associated with the pairs of vertices (B,C) and (C, A): Corollary 5. The symmedian point K of any triangle ABC is located in the common part of the interiors of the three ellipses with foci A and B, with foci B and C, with foci C and A, and with respective major axes 2AB, 2BC, 2CA (boundary included). It is known that K is always interior to the circle with diameter GH where G is the centroid and H the orthocenter (see []). However, the common part interior to the three ellipses can be much smaller (Figure ). Figure Lastly, we deduce a nice geometric inequality, perhaps difficult to prove directly: Corollary 6. Let m a,m b,m c be the lengths of the medians of a triangle ABC with sides a = BC,b = CA,c = AB. Then, the following inequality holds: max{bm c + cm b,cm a + am c,am b + bm a } a2 + b 2 + c 2. (2) Interestingly, substituting m a,m b,m c, a 4, b 4, c 4 for a, b, c, m a,m b,m c, respectively, leaves the inequality unchanged, meaning that the inequality is its own
5 Constructing a triangle from two vertices and the symmedian point 1 median-dual ([5] p. 109). No hope then to derive it from a known inequality through median duality! Proof. We know that the symmedian point K of ΔABC is the center of masses of (A, a 2 ), (B,b 2 ), (C, c 2 ), that is, (a 2 + b 2 + c 2 )K = a 2 A + b 2 B + c 2 C. It follows that (a 2 + b 2 + c 2 ) AK = b 2 AB + c 2 AC and so (a 2 + b 2 + c 2 ) 2 AK 2 = b 4 c 2 + c 4 b 2 +(b 2 c 2 )(2 AB AC) = b 2 c 2 (b 2 + c 2 + b 2 + c 2 a 2 ) = 4b 2 c 2 m 2 a. As a result, KA = 2bcma and similarly we obtain KB = 2cam a 2 +b 2 +c 2 b. The a 2 +b 2 +c 2 inequality KA + KB 2c now rewrites as bm a + am b a2 +b 2 +c 2. Cyclically, the numbers bm c +cm b and cm a +am c are also less than or equal to a2 +b 2 +c 2. Note that equality holds if and only if ΔABC is a root-mean-square triangle, that is, if and only if one of the relations 2a 2 = b 2 + c 2, 2b 2 = c 2 + a 2, 2c 2 = a 2 + b 2 is satisfied. References [1] M. Bataille, Characterizing a symmedian, Crux Math., 4 (2017) 148. [2] M. Bataille, On the centres of root-mean-square triangles, Crux Math., 44 (2018) [] C. J. Bradley and G. C. Smith, The locations of triangle centers, Forum Geom., 6 (2006) [4] J. C. Fisher, Recurring Crux configurations, Crux Math., 7 (2011) [5] D. S. Mintrinović, J. E. Pečarić, and V. Volenec, Recent Advances in Geometric Inequalities, Kluwer, [6] P. Schreck, P. Mathis, V. Marinković, and P. Janicić, Wernick s List: A Final Update, Forum Geom., 16 (2016) [7] W. Wernick, Triangle Constructions with Three Located Points, Math. Mag., 55 (1982) Michel Bataille: 6 square des Boulots, Franqueville-Saint-Pierre, France address: michelbataille@wanadoo.fr
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE
Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.6, (207), Issue, pp.45-57 A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE PAUL YIU AND XIAO-DONG ZHANG
More informationSome Collinearities in the Heptagonal Triangle
Forum Geometricorum Volume 16 (2016) 249 256. FRUM GEM ISSN 1534-1178 Some ollinearities in the Heptagonal Triangle bdilkadir ltintaş bstract. With the methods of barycentric coordinates, we establish
More informationAnother Proof of van Lamoen s Theorem and Its Converse
Forum Geometricorum Volume 5 (2005) 127 132. FORUM GEOM ISSN 1534-1178 Another Proof of van Lamoen s Theorem and Its Converse Nguyen Minh Ha Abstract. We give a proof of Floor van Lamoen s theorem and
More informationThe Apollonius Circle and Related Triangle Centers
Forum Geometricorum Qolume 3 (2003) 187 195. FORUM GEOM ISSN 1534-1178 The Apollonius Circle and Related Triangle Centers Milorad R. Stevanović Abstract. We give a simple construction of the Apollonius
More informationchapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true?
chapter vector geometry solutions V. Exercise A. For the shape shown, find a single vector which is equal to a)!!! " AB + BC AC b)! AD!!! " + DB AB c)! AC + CD AD d)! BC + CD!!! " + DA BA e) CD!!! " "
More informationOn the Circumcenters of Cevasix Configurations
Forum Geometricorum Volume 3 (2003) 57 63. FORUM GEOM ISSN 1534-1178 On the ircumcenters of evasix onfigurations lexei Myakishev and Peter Y. Woo bstract. We strengthen Floor van Lamoen s theorem that
More informationBounds for Elements of a Triangle Expressed by R, r, and s
Forum Geometricorum Volume 5 05) 99 03. FOUM GEOM ISSN 534-78 Bounds for Elements of a Triangle Expressed by, r, and s Temistocle Bîrsan Abstract. Assume that a triangle is defined by the triple, r, s)
More informationMidterm Review Packet. Geometry: Midterm Multiple Choice Practice
: Midterm Multiple Choice Practice 1. In the diagram below, a square is graphed in the coordinate plane. A reflection over which line does not carry the square onto itself? (1) (2) (3) (4) 2. A sequence
More informationThe Droz-Farny Circles of a Convex Quadrilateral
Forum Geometricorum Volume 11 (2011) 109 119. FORUM GEOM ISSN 1534-1178 The Droz-Farny Circles of a Convex Quadrilateral Maria Flavia Mammana, Biagio Micale, and Mario Pennisi Abstract. The Droz-Farny
More informationOn Emelyanov s Circle Theorem
Journal for Geometry and Graphics Volume 9 005, No., 55 67. On Emelyanov s ircle Theorem Paul Yiu Department of Mathematical Sciences, Florida Atlantic University Boca Raton, Florida, 3343, USA email:
More information2013 Sharygin Geometry Olympiad
Sharygin Geometry Olympiad 2013 First Round 1 Let ABC be an isosceles triangle with AB = BC. Point E lies on the side AB, and ED is the perpendicular from E to BC. It is known that AE = DE. Find DAC. 2
More informationClassical Theorems in Plane Geometry 1
BERKELEY MATH CIRCLE 1999 2000 Classical Theorems in Plane Geometry 1 Zvezdelina Stankova-Frenkel UC Berkeley and Mills College Note: All objects in this handout are planar - i.e. they lie in the usual
More informationThe Apollonius Circle as a Tucker Circle
Forum Geometricorum Volume 2 (2002) 175 182 FORUM GEOM ISSN 1534-1178 The Apollonius Circle as a Tucker Circle Darij Grinberg and Paul Yiu Abstract We give a simple construction of the circular hull of
More informationChapter 6. Basic triangle centers. 6.1 The Euler line The centroid
hapter 6 asic triangle centers 6.1 The Euler line 6.1.1 The centroid Let E and F be the midpoints of and respectively, and G the intersection of the medians E and F. onstruct the parallel through to E,
More informationProblems First day. 8 grade. Problems First day. 8 grade
First day. 8 grade 8.1. Let ABCD be a cyclic quadrilateral with AB = = BC and AD = CD. ApointM lies on the minor arc CD of its circumcircle. The lines BM and CD meet at point P, thelinesam and BD meet
More information1/19 Warm Up Fast answers!
1/19 Warm Up Fast answers! The altitudes are concurrent at the? Orthocenter The medians are concurrent at the? Centroid The perpendicular bisectors are concurrent at the? Circumcenter The angle bisectors
More informationQ1. If (1, 2) lies on the circle. x 2 + y 2 + 2gx + 2fy + c = 0. which is concentric with the circle x 2 + y 2 +4x + 2y 5 = 0 then c =
Q1. If (1, 2) lies on the circle x 2 + y 2 + 2gx + 2fy + c = 0 which is concentric with the circle x 2 + y 2 +4x + 2y 5 = 0 then c = a) 11 b) -13 c) 24 d) 100 Solution: Any circle concentric with x 2 +
More informationINVERSION IN THE PLANE BERKELEY MATH CIRCLE
INVERSION IN THE PLANE BERKELEY MATH CIRCLE ZVEZDELINA STANKOVA MILLS COLLEGE/UC BERKELEY SEPTEMBER 26TH 2004 Contents 1. Definition of Inversion in the Plane 1 Properties of Inversion 2 Problems 2 2.
More informationBerkeley Math Circle, May
Berkeley Math Circle, May 1-7 2000 COMPLEX NUMBERS IN GEOMETRY ZVEZDELINA STANKOVA FRENKEL, MILLS COLLEGE 1. Let O be a point in the plane of ABC. Points A 1, B 1, C 1 are the images of A, B, C under symmetry
More informationAdditional Mathematics Lines and circles
Additional Mathematics Lines and circles Topic assessment 1 The points A and B have coordinates ( ) and (4 respectively. Calculate (i) The gradient of the line AB [1] The length of the line AB [] (iii)
More informationDistances Among the Feuerbach Points
Forum Geometricorum Volume 16 016) 373 379 FORUM GEOM ISSN 153-1178 Distances mong the Feuerbach Points Sándor Nagydobai Kiss bstract We find simple formulas for the distances from the Feuerbach points
More informationVectors - Applications to Problem Solving
BERKELEY MATH CIRCLE 00-003 Vectors - Applications to Problem Solving Zvezdelina Stankova Mills College& UC Berkeley 1. Well-known Facts (1) Let A 1 and B 1 be the midpoints of the sides BC and AC of ABC.
More informationTHE CYCLIC QUADRANGLE IN THE ISOTROPIC PLANE
SARAJEVO JOURNAL OF MATHEMATICS Vol.7 (0) (011), 65 75 THE CYCLIC QUADRANGLE IN THE ISOTROPIC PLANE V. VOLENEC, J. BEBAN-BRKIĆ AND M. ŠIMIĆ Abstract. In [15], [] we focused on the geometry of the non-tangential
More informationOn an Erdős Inscribed Triangle Inequality
Forum Geometricorum Volume 5 (005) 137 141. FORUM GEOM ISSN 1534-1178 On an Erdős Inscribed Triangle Inequality Ricardo M. Torrejón Abstract. A comparison between the area of a triangle that of an inscribed
More information0112ge. Geometry Regents Exam Line n intersects lines l and m, forming the angles shown in the diagram below.
Geometry Regents Exam 011 011ge 1 Line n intersects lines l and m, forming the angles shown in the diagram below. 4 In the diagram below, MATH is a rhombus with diagonals AH and MT. Which value of x would
More informationBicevian Tucker Circles
Forum Geometricorum Volume 7 (2007) 87 97. FORUM GEOM ISSN 1534-1178 icevian Tucker ircles ernard Gibert bstract. We prove that there are exactly ten bicevian Tucker circles and show several curves containing
More informationHarmonic quadrangle in isotropic plane
Turkish Journal of Mathematics http:// journals. tubitak. gov. tr/ math/ Research Article Turk J Math (018) 4: 666 678 c TÜBİTAK doi:10.3906/mat-1607-35 Harmonic quadrangle in isotropic plane Ema JURKIN
More informationThe Arbelos and Nine-Point Circles
Forum Geometricorum Volume 7 (2007) 115 120. FORU GEO ISSN 1534-1178 The rbelos and Nine-Point Circles uang Tuan ui bstract. We construct some new rchimedean circles in an arbelos in connection with the
More informationarxiv: v1 [math.ho] 29 Nov 2017
The Two Incenters of the Arbitrary Convex Quadrilateral Nikolaos Dergiades and Dimitris M. Christodoulou ABSTRACT arxiv:1712.02207v1 [math.ho] 29 Nov 2017 For an arbitrary convex quadrilateral ABCD with
More informationHeptagonal Triangles and Their Companions
Forum Geometricorum Volume 9 (009) 15 148. FRUM GEM ISSN 1534-1178 Heptagonal Triangles and Their ompanions Paul Yiu bstract. heptagonal triangle is a non-isosceles triangle formed by three vertices of
More informationNozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch
Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. in the parallelogram, each two opposite
More information0811ge. Geometry Regents Exam
0811ge 1 The statement "x is a multiple of 3, and x is an even integer" is true when x is equal to 1) 9 ) 8 3) 3 4) 6 In the diagram below, ABC XYZ. 4 Pentagon PQRST has PQ parallel to TS. After a translation
More informationConstruction of Ajima Circles via Centers of Similitude
Forum Geometricorum Volume 15 (2015) 203 209. FORU GEO SS 1534-1178 onstruction of jima ircles via enters of Similitude ikolaos Dergiades bstract. We use the notion of the centers of similitude of two
More informationIsogonal Conjugacy Through a Fixed Point Theorem
Forum Geometricorum Volume 16 (016 171 178. FORUM GEOM ISSN 1534-1178 Isogonal onjugacy Through a Fixed Point Theorem Sándor Nagydobai Kiss and Zoltán Kovács bstract. For an arbitrary point in the plane
More informationGeometry in the Complex Plane
Geometry in the Complex Plane Hongyi Chen UNC Awards Banquet 016 All Geometry is Algebra Many geometry problems can be solved using a purely algebraic approach - by placing the geometric diagram on a coordinate
More information1 Line n intersects lines l and m, forming the angles shown in the diagram below. 4 In the diagram below, MATH is a rhombus with diagonals AH and MT.
1 Line n intersects lines l and m, forming the angles shown in the diagram below. 4 In the diagram below, MATH is a rhombus with diagonals AH and MT. Which value of x would prove l m? 1) 2.5 2) 4.5 3)
More informationComplex numbers in the realm of Euclidean Geometry
Complex numbers in the realm of Euclidean Geometry Finbarr Holland February 7, 014 1 Introduction Before discussing the complex forms of lines and circles, we recall some familiar facts about complex numbers.
More informationNagel, Speiker, Napoleon, Torricelli. Centroid. Circumcenter 10/6/2011. MA 341 Topics in Geometry Lecture 17
Nagel, Speiker, Napoleon, Torricelli MA 341 Topics in Geometry Lecture 17 Centroid The point of concurrency of the three medians. 07-Oct-2011 MA 341 2 Circumcenter Point of concurrency of the three perpendicular
More informationTriangles III. Stewart s Theorem, Orthocenter, Euler Line. 23-Sept-2011 MA
Triangles III Stewart s Theorem, Orthocenter, Euler Line 23-Sept-2011 MA 341 001 1 Stewart s Theorem (1746) With the measurements given in the triangle below, the following relationship holds: a 2 n +
More informationThree Natural Homoteties of The Nine-Point Circle
Forum Geometricorum Volume 13 (2013) 209 218. FRUM GEM ISS 1534-1178 Three atural omoteties of The ine-point ircle Mehmet Efe kengin, Zeyd Yusuf Köroğlu, and Yiğit Yargiç bstract. Given a triangle with
More informationA Sequence of Triangles and Geometric Inequalities
Forum Geometricorum Volume 9 (009) 91 95. FORUM GEOM ISSN 1534-1178 A Sequence of Triangles and Geometric Inequalities Dan Marinescu, Mihai Monea, Mihai Opincariu, and Marian Stroe Abstract. We construct
More informationA Note on the Barycentric Square Roots of Kiepert Perspectors
Forum Geometricorum Volume 6 (2006) 263 268. FORUM GEOM ISSN 1534-1178 Note on the arycentric Square Roots of Kiepert erspectors Khoa Lu Nguyen bstract. Let be an interior point of a given triangle. We
More informationPairs of Cocentroidal Inscribed and Circumscribed Triangles
Forum Geometricorum Volume 15 (2015) 185 190. FORUM GEOM ISSN 1534-1178 Pairs of Cocentroidal Inscribed and Circumscribed Triangles Gotthard Weise Abstract. Let Δ be a reference triangle and P a point
More informationThe Kiepert Pencil of Kiepert Hyperbolas
Forum Geometricorum Volume 1 (2001) 125 132. FORUM GEOM ISSN 1534-1178 The Kiepert Pencil of Kiepert Hyperbolas Floor van Lamoen and Paul Yiu bstract. We study Kiepert triangles K(φ) and their iterations
More informationLemoine Circles Radius Calculus
Lemoine Circles Radius Calculus Ion Pătrașcu, professor, Frații Buzești National College,Craiova, Romania Florentin Smarandache, professor, New Mexico University, U.S.A. For the calculus of the first Lemoine
More informationGeometry Problem booklet
Geometry Problem booklet Assoc. Prof. Cornel Pintea E-mail: cpintea math.ubbcluj.ro Contents 1 Week 1: Vector algebra 1 1.1 Brief theoretical background. Free vectors..................... 1 1.1.1 Operations
More informationExamples: Identify three pairs of parallel segments in the diagram. 1. AB 2. BC 3. AC. Write an equation to model this theorem based on the figure.
5.1: Midsegments of Triangles NOTE: Midsegments are also to the third side in the triangle. Example: Identify the 3 midsegments in the diagram. Examples: Identify three pairs of parallel segments in the
More informationThe Lemoine Cubic and Its Generalizations
Forum Geometricorum Volume 2 (2002) 47 63. FRUM GEM ISSN 1534-1178 The Lemoine ubic and Its Generalizations ernard Gibert bstract. For a given triangle, the Lemoine cubic is the locus of points whose cevian
More informationComplex Numbers in Geometry
Complex Numers in Geometry Seastian Jeon Decemer 3, 206 The Complex Plane. Definitions I assume familiarity with most, if not all, of the following definitions. Some knowledge of linear algera is also
More informationLOCUS. Definition: The set of all points (and only those points) which satisfy the given geometrical condition(s) (or properties) is called a locus.
LOCUS Definition: The set of all points (and only those points) which satisfy the given geometrical condition(s) (or properties) is called a locus. Eg. The set of points in a plane which are at a constant
More information2. In ABC, the measure of angle B is twice the measure of angle A. Angle C measures three times the measure of angle A. If AC = 26, find AB.
2009 FGCU Mathematics Competition. Geometry Individual Test 1. You want to prove that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex. Which postulate/theorem
More informationXX Asian Pacific Mathematics Olympiad
XX Asian Pacific Mathematics Olympiad March, 008 Problem 1. Let ABC be a triangle with A < 60. Let X and Y be the points on the sides AB and AC, respectively, such that CA + AX = CB + BX and BA + AY =
More informationA Note on Reflections
Forum Geometricorum Volume 14 (2014) 155 161. FORUM GEOM SSN 1534-1178 Note on Reflections Emmanuel ntonio José García bstract. We prove some simple results associated with the triangle formed by the reflections
More informationStatics and the Moduli Space of Triangles
Forum Geometricorum Volume 5 (2005) 181 10. FORUM GEOM ISSN 1534-1178 Statics and the Moduli Space of Triangles Geoff C. Smith Abstract. The variance of a weighted collection of points is used to prove
More informationIsogonal Conjugates. Navneel Singhal October 9, Abstract
Isogonal Conjugates Navneel Singhal navneel.singhal@ymail.com October 9, 2016 Abstract This is a short note on isogonality, intended to exhibit the uses of isogonality in mathematical olympiads. Contents
More informationAnother Variation on the Steiner-Lehmus Theme
Forum Geometricorum Volume 8 (2008) 131 140. FORUM GOM ISSN 1534-1178 Another Variation on the Steiner-Lehmus Theme Sadi Abu-Saymeh, Mowaffaq Hajja, and Hassan Ali ShahAli Abstract. Let the internal angle
More informationChapter 2. The laws of sines and cosines. 2.1 The law of sines. Theorem 2.1 (The law of sines). Let R denote the circumradius of a triangle ABC.
hapter 2 The laws of sines and cosines 2.1 The law of sines Theorem 2.1 (The law of sines). Let R denote the circumradius of a triangle. 2R = a sin α = b sin β = c sin γ. α O O α as Since the area of a
More informationForum Geometricorum Volume 13 (2013) 1 6. FORUM GEOM ISSN Soddyian Triangles. Frank M. Jackson
Forum Geometricorum Volume 3 (203) 6. FORUM GEOM ISSN 534-78 Soddyian Triangles Frank M. Jackson bstract. Soddyian triangle is a triangle whose outer Soddy circle has degenerated into a straight line.
More informationMathematics Revision Guides Vectors Page 1 of 19 Author: Mark Kudlowski M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier VECTORS
Mathematics Revision Guides Vectors Page of 9 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier VECTORS Version:.4 Date: 05-0-05 Mathematics Revision Guides Vectors Page of 9 VECTORS
More information(D) (A) Q.3 To which of the following circles, the line y x + 3 = 0 is normal at the point ? 2 (A) 2
CIRCLE [STRAIGHT OBJECTIVE TYPE] Q. The line x y + = 0 is tangent to the circle at the point (, 5) and the centre of the circles lies on x y = 4. The radius of the circle is (A) 3 5 (B) 5 3 (C) 5 (D) 5
More informationSOME NEW INEQUALITIES FOR AN INTERIOR POINT OF A TRIANGLE JIAN LIU. 1. Introduction
Journal of Mathematical Inequalities Volume 6, Number 2 (2012), 195 204 doi:10.7153/jmi-06-20 OME NEW INEQUALITIE FOR AN INTERIOR POINT OF A TRIANGLE JIAN LIU (Communicated by. egura Gomis) Abstract. In
More information0114ge. Geometry Regents Exam 0114
0114ge 1 The midpoint of AB is M(4, 2). If the coordinates of A are (6, 4), what are the coordinates of B? 1) (1, 3) 2) (2, 8) 3) (5, 1) 4) (14, 0) 2 Which diagram shows the construction of a 45 angle?
More informationThe Napoleon Configuration
Forum eometricorum Volume 2 (2002) 39 46. FORUM EOM ISSN 1534-1178 The Napoleon onfiguration illes outte bstract. It is an elementary fact in triangle geometry that the two Napoleon triangles are equilateral
More informationSenior Math Circles February 18, 2009 Conics III
University of Waterloo Faculty of Mathematics Senior Math Circles February 18, 2009 Conics III Centre for Education in Mathematics and Computing Eccentricity of Conics Fix a point F called the focus, a
More informationQUESTION BANK ON STRAIGHT LINE AND CIRCLE
QUESTION BANK ON STRAIGHT LINE AND CIRCLE Select the correct alternative : (Only one is correct) Q. If the lines x + y + = 0 ; 4x + y + 4 = 0 and x + αy + β = 0, where α + β =, are concurrent then α =,
More informationSTRAIGHT LINES EXERCISE - 3
STRAIGHT LINES EXERCISE - 3 Q. D C (3,4) E A(, ) Mid point of A, C is B 3 E, Point D rotation of point C(3, 4) by angle 90 o about E. 3 o 3 3 i4 cis90 i 5i 3 i i 5 i 5 D, point E mid point of B & D. So
More informationMathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions
Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions Quiz #1. Tuesday, 17 January, 2012. [10 minutes] 1. Given a line segment AB, use (some of) Postulates I V,
More informationCO-ORDINATE GEOMETRY. 1. Find the points on the y axis whose distances from the points (6, 7) and (4,-3) are in the. ratio 1:2.
UNIT- CO-ORDINATE GEOMETRY Mathematics is the tool specially suited for dealing with abstract concepts of any ind and there is no limit to its power in this field.. Find the points on the y axis whose
More information1 Solution of Final. Dr. Franz Rothe December 25, Figure 1: Dissection proof of the Pythagorean theorem in a special case
Math 3181 Dr. Franz Rothe December 25, 2012 Name: 1 Solution of Final Figure 1: Dissection proof of the Pythagorean theorem in a special case 10 Problem 1. Given is a right triangle ABC with angle α =
More informationTwo applications of the theorem of Carnot
Annales Mathematicae et Informaticae 40 (2012) pp. 135 144 http://ami.ektf.hu Two applications of the theorem of Carnot Zoltán Szilasi Institute of Mathematics, MTA-DE Research Group Equations, Functions
More informationArticle 13. Ex-points and eight intersecting circles
Article 13 Ex-points and eight intersecting circles Christopher J Bradley H' F P3 W' L' A V' N E P2 M U O Q V P J H W M' B R C N' L To T S U' P1 D Figure 1. Introduction 1 In this article we use areal
More informationCh 5 Practice Exam. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.
Name: Class: Date: Ch 5 Practice Exam Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Find the value of x. The diagram is not to scale. a. 32 b. 50 c.
More informationA sequence of thoughts on constructible angles.
A sequence of thoughts on constructible angles. Dan Franklin & Kevin Pawski Department of Mathematics, SUNY New Paltz, New Paltz, NY 12561 Nov 23, 2002 1 Introduction In classical number theory the algebraic
More informationAnalytic Geometry MAT 1035
Analytic Geometry MAT 035 5.09.04 WEEKLY PROGRAM - The first week of the semester, we will introduce the course and given a brief outline. We continue with vectors in R n and some operations including
More informationThe Distance Formula. The Midpoint Formula
Math 120 Intermediate Algebra Sec 9.1: Distance Midpoint Formulas The Distance Formula The distance between two points P 1 = (x 1, y 1 ) P 2 = (x 1, y 1 ), denoted by d(p 1, P 2 ), is d(p 1, P 2 ) = (x
More informationIon Patrascu, Florentin Smarandache Theorems with Parallels Taken through a Triangle s Vertices and Constructions Performed only with the Ruler
Theorems with Parallels Taken through a Triangle s Vertices and Constructions Performed only with the Ruler In Ion Patrascu, Florentin Smarandache: Complements to Classic Topics of Circles Geometry. Brussels
More information0809ge. Geometry Regents Exam Based on the diagram below, which statement is true?
0809ge 1 Based on the diagram below, which statement is true? 3 In the diagram of ABC below, AB AC. The measure of B is 40. 1) a b ) a c 3) b c 4) d e What is the measure of A? 1) 40 ) 50 3) 70 4) 100
More informationCevian Projections of Inscribed Triangles and Generalized Wallace Lines
Forum Geometricorum Volume 16 (2016) 241 248. FORUM GEOM ISSN 1534-1178 Cevian Projections of Inscribed Triangles and Generalized Wallace Lines Gotthard Weise 1. Notations Abstract. Let Δ=ABC be a reference
More informationComments about Chapters 4 and 5 of the Math 5335 (Geometry I) text Joel Roberts November 5, 2003; revised October 18, 2004 and October 2005
Comments about Chapters 4 and 5 of the Math 5335 (Geometry I) text Joel Roberts November 5, 2003; revised October 8, 2004 and October 2005 Contents: Heron's formula (Theorem 9 in 4.5). 4.4: Another proof
More informationSMT Power Round Solutions : Poles and Polars
SMT Power Round Solutions : Poles and Polars February 18, 011 1 Definition and Basic Properties 1 Note that the unit circles are not necessary in the solutions. They just make the graphs look nicer. (1).0
More informationEdexcel New GCE A Level Maths workbook Circle.
Edexcel New GCE A Level Maths workbook Circle. Edited by: K V Kumaran kumarmaths.weebly.com 1 Finding the Midpoint of a Line To work out the midpoint of line we need to find the halfway point Midpoint
More informationIsotomic Inscribed Triangles and Their Residuals
Forum Geometricorum Volume 3 (2003) 125 134. FORUM GEOM ISSN 1534-1178 Isotomic Inscribed Triangles and Their Residuals Mario Dalcín bstract. We prove some interesting results on inscribed triangles which
More informationTwo applications of the theorem of Carnot
Two applications of the theorem of Carnot Zoltán Szilasi Abstract Using the theorem of Carnot we give elementary proofs of two statements of C Bradley We prove his conjecture concerning the tangents to
More informationMathematics 2260H Geometry I: Euclidean geometry Trent University, Fall 2016 Solutions to the Quizzes
Mathematics 2260H Geometry I: Euclidean geometry Trent University, Fall 2016 Solutions to the Quizzes Quiz #1. Wednesday, 13 September. [10 minutes] 1. Suppose you are given a line (segment) AB. Using
More informationGeometry JWR. Monday September 29, 2003
Geometry JWR Monday September 29, 2003 1 Foundations In this section we see how to view geometry as algebra. The ideas here should be familiar to the reader who has learned some analytic geometry (including
More informationAnalytic Geometry MAT 1035
Analytic Geometry MAT 035 5.09.04 WEEKLY PROGRAM - The first week of the semester, we will introduce the course and given a brief outline. We continue with vectors in R n and some operations including
More informationDefinitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg ( )
Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg (2005-02-16) Logic Rules (Greenberg): Logic Rule 1 Allowable justifications.
More informationSimple Proofs of Feuerbach s Theorem and Emelyanov s Theorem
Forum Geometricorum Volume 18 (2018) 353 359. FORUM GEOM ISSN 1534-1178 Simple Proofs of Feuerbach s Theorem and Emelyanov s Theorem Nikolaos Dergiades and Tran Quang Hung Abstract. We give simple proofs
More informationGeneralized Mandart Conics
Forum Geometricorum Volume 4 (2004) 177 198. FORUM GEOM ISSN 1534-1178 Generalized Mandart onics ernard Gibert bstract. We consider interesting conics associated with the configuration of three points
More informationTOPIC 4 Line and Angle Relationships. Good Luck To. DIRECTIONS: Answer each question and show all work in the space provided.
Good Luck To Period Date DIRECTIONS: Answer each question and show all work in the space provided. 1. Name a pair of corresponding angles. 1 3 2 4 5 6 7 8 A. 1 and 4 C. 2 and 7 B. 1 and 5 D. 2 and 4 2.
More informationThe Locations of the Brocard Points
Forum Geometricorum Volume 6 (2006) 71 77. FORUM GEOM ISSN 1534-1178 The Locations of the Brocard Points Christopher J. Bradley and Geoff C. Smith Abstract. We fix and scale to constant size the orthocentroidal
More informationIsogonal Conjugates in a Tetrahedron
Forum Geometricorum Volume 16 (2016) 43 50. FORUM GEOM ISSN 1534-1178 Isogonal Conjugates in a Tetrahedron Jawad Sadek, Majid Bani-Yaghoub, and Noah H. Rhee Abstract. The symmedian point of a tetrahedron
More informationP1 Chapter 6 :: Circles
P1 Chapter 6 :: Circles jfrost@tiffin.kingston.sch.uk www.drfrostmaths.com @DrFrostMaths Last modified: 11 th August 2017 Use of DrFrostMaths for practice Register for free at: www.drfrostmaths.com/homework
More informationUse this space for computations. 1 In trapezoid RSTV below with bases RS and VT, diagonals RT and SV intersect at Q.
Part I Answer all 28 questions in this part. Each correct answer will receive 2 credits. For each statement or question, choose the word or expression that, of those given, best completes the statement
More informationGeometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014
Geometry lass Examples (July 1) Paul Yiu Department of Mathematics Florida tlantic University c b a Summer 2014 1 Example 1(a). Given a triangle, the intersection P of the perpendicular bisector of and
More informationSample Question Paper Mathematics First Term (SA - I) Class IX. Time: 3 to 3 ½ hours
Sample Question Paper Mathematics First Term (SA - I) Class IX Time: 3 to 3 ½ hours M.M.:90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided
More informationHomework Assignments Math /02 Fall 2017
Homework Assignments Math 119-01/02 Fall 2017 Assignment 1 Due date : Wednesday, August 30 Section 6.1, Page 178: #1, 2, 3, 4, 5, 6. Section 6.2, Page 185: #1, 2, 3, 5, 6, 8, 10-14, 16, 17, 18, 20, 22,
More informationConcurrency and Collinearity
Concurrency and Collinearity Victoria Krakovna vkrakovna@gmail.com 1 Elementary Tools Here are some tips for concurrency and collinearity questions: 1. You can often restate a concurrency question as a
More informationNINETEENTH IRISH MATHEMATICAL OLYMPIAD. Saturday, 6 May a.m. 1 p.m. First Paper. 1. Are there integers x, y, and z which satisfy the equation
NINETEENTH IRISH MATHEMATICAL OLYMPIAD Saturday, 6 May 2006 10 a.m. 1 p.m. First Paper 1. Are there integers x, y, and z which satisfy the equation when (a) n = 2006 (b) n = 2007? z 2 = (x 2 + 1)(y 2 1)
More informationTheorem on altitudes and the Jacobi identity
Theorem on altitudes and the Jacobi identity A. Zaslavskiy and M. Skopenkov Solutions. First let us give a table containing the answers to all the problems: Algebraic object Geometric sense A apointa a
More information